Let I ( x ) = 6 sin 2 x ( 1 - cot x ) 2 d x . If I ( 0 ) = 3 , then is equal to [2024]

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Solution

Q.
Let $I (x) = \int \frac{6}{\sin^{2} x {(1 - \cot x)}^{2}} d x .$ If $I (0) = 3,$ then $I (\frac{π}{12})$ is equal to [2024]

1 $6 \sqrt{3}$

2 $3 \sqrt{3}$

3 $\sqrt{3}$

4 $2 \sqrt{3}$

Ans.
(2)

$I (x) = \int \frac{6}{\sin^{2} x {(1 - \cot x)}^{2}} d x = \int \frac{6 c o s e c^{2} x}{{(1 - \cot x)}^{2}} d x$

Put $1 - \cot x = t \Rightarrow c o s e c^{2} x d x = d t$

$\Rightarrow I = \int \frac{6 d t}{t^{2}} = \frac{- 6}{t} + C \Rightarrow I = \frac{- 6}{1 - \cot x} + C$

Now, $I (0) = 3 \Rightarrow 3 = C$

$\Rightarrow I (x) = 3 - \frac{6}{1 - \cot x} \Rightarrow I (\frac{π}{12}) = 3 - \frac{6}{1 - \cot \frac{π}{12}}$

$= 3 - \frac{6}{1 - (2 + \sqrt{3})} = \frac{- 3 - 3 \sqrt{3} - 6}{- 1 - \sqrt{3}} = \frac{- 9 - 3 \sqrt{3}}{- 1 - \sqrt{3}}$

$= \frac{(9 + 3 \sqrt{3}) (1 - \sqrt{3})}{- 2} = \frac{9 - 9 \sqrt{3} + 3 \sqrt{3} - 9}{- 2} = 3 \sqrt{3}$

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